Find points at a given distance on a line of given slope
Given the co-ordinates of a 2-dimensional point p(x0, y0). Find the points at a distance L away from it, such that the line formed by joining these points has a slope of M.
Examples:
Input : p = (2, 1) L = sqrt(2) M = 1 Output :3, 2 1, 0 Explanation: The two points are sqrt(2) distance away from the source and have the required slope m = 1. Input : p = (1, 0) L = 5 M = 0 Output : 6, 0 -4, 0
We need to find two points that are L distance from given point, on a line with slope M.
The idea has been introduced in below post.
Find Corners of Rectangle using mid points
Based on the input slope, the problem can be classified into 3 categories.
- If slope is zero, we just need to adjust the x coordinate of the source point
- If slope is infinite, the we need to adjust the y coordinate
- For other values of slope, we can use the following equations to find the points
Now using the above formula we can find the required points.
C++
// C++ program to find the points on a line of // slope M at distance L #include <bits/stdc++.h> using namespace std; // structure to represent a co-ordinate // point struct Point { float x, y; Point() { x = y = 0; } Point( float a, float b) { x = a, y = b; } }; // Function to print pair of points at // distance 'l' and having a slope 'm' // from the source void printPoints(Point source, float l, int m) { // m is the slope of line, and the // required Point lies distance l // away from the source Point Point a, b; // slope is 0 if (m == 0) { a.x = source.x + l; a.y = source.y; b.x = source.x - l; b.y = source.y; } // if slope is infinite else if (m == std::numeric_limits< float > ::max()) { a.x = source.x; a.y = source.y + l; b.x = source.x; b.y = source.y - l; } else { float dx = (l / sqrt (1 + (m * m))); float dy = m * dx; a.x = source.x + dx; a.y = source.y + dy; b.x = source.x - dx; b.y = source.y - dy; } // print the first Point cout << a.x << ", " << a.y << endl; // print the second Point cout << b.x << ", " << b.y << endl; } // driver function int main() { Point p(2, 1), q(1, 0); printPoints(p, sqrt (2), 1); cout << endl; printPoints(q, 5, 0); return 0; } |
Java
// Java program to find the points on // a line of slope M at distance L class GFG{ // Class to represent a co-ordinate // point static class Point { float x, y; Point() { x = y = 0 ; } Point( float a, float b) { x = a; y = b; } }; // Function to print pair of points at // distance 'l' and having a slope 'm' // from the source static void printPoints(Point source, float l, int m) { // m is the slope of line, and the // required Point lies distance l // away from the source Point Point a = new Point(); Point b = new Point(); // Slope is 0 if (m == 0 ) { a.x = source.x + l; a.y = source.y; b.x = source.x - l; b.y = source.y; } // If slope is infinite else if (Double.isInfinite(m)) { a.x = source.x; a.y = source.y + l; b.x = source.x; b.y = source.y - l; } else { float dx = ( float )(l / Math.sqrt( 1 + (m * m))); float dy = m * dx; a.x = source.x + dx; a.y = source.y + dy; b.x = source.x - dx; b.y = source.y - dy; } // Print the first Point System.out.println(a.x + ", " + a.y); // Print the second Point System.out.println(b.x + ", " + b.y); } // Driver code public static void main(String[] args) { Point p = new Point( 2 , 1 ), q = new Point( 1 , 0 ); printPoints(p, ( float )Math.sqrt( 2 ), 1 ); System.out.println(); printPoints(q, 5 , 0 ); } } // This code is contributed by Rajnis09 |
C#
// C# program to find the points on // a line of slope M at distance L using System; class GFG{ // Class to represent a co-ordinate // point public class Point { public float x, y; public Point() { x = y = 0; } public Point( float a, float b) { x = a; y = b; } }; // Function to print pair of points at // distance 'l' and having a slope 'm' // from the source static void printPoints(Point source, float l, int m) { // m is the slope of line, and the // required Point lies distance l // away from the source Point Point a = new Point(); Point b = new Point(); // Slope is 0 if (m == 0) { a.x = source.x + l; a.y = source.y; b.x = source.x - l; b.y = source.y; } // If slope is infinite else if (Double.IsInfinity(m)) { a.x = source.x; a.y = source.y + l; b.x = source.x; b.y = source.y - l; } else { float dx = ( float )(l / Math.Sqrt( 1 + (m * m))); float dy = m * dx; a.x = source.x + dx; a.y = source.y + dy; b.x = source.x - dx; b.y = source.y - dy; } // Print the first Point Console.WriteLine(a.x + ", " + a.y); // Print the second Point Console.WriteLine(b.x + ", " + b.y); } // Driver code public static void Main(String[] args) { Point p = new Point(2, 1), q = new Point(1, 0); printPoints(p, ( float )Math.Sqrt(2), 1); Console.WriteLine(); printPoints(q, 5, 0); } } // This code is contributed by Amit Katiyar |
Output:
3, 2 1, 0 6, 0 -4, 0
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